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Suppose that $F:S^{n-1}\to A$ is a map of sets from the unit sphere in $\mathbb R^n$ to an abelian group, and that the sum $F(v_1)+\dots +F(v_n)$ over an orthonormal basis is independent of the basis. Does it follow that $F$ is a constant function?

This is clearly false for $n=2$. I am wondering if it is true for sufficiently large $n$.

ADDED LATER The $\mathbb R$-valued examples in Cranch's answer may be combined into a single example $v\mapsto v\otimes v$ with values in $\mathbb R^n\otimes \mathbb R^n$, or $n\times n$ matrices. Its image generates the group of symmetric matrices with integer trace. It seems reasonable to expect that every continuous real-valued example comes from this one -- in other words has the form $v\mapsto B(v,v)$ for symmetric bilinear $B$. Maybe this can be worked out using Sawin's suggestion about representations of $O(n)$. But I was also curious about the general case, where the target group might not be (uniquely) divisible.

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    $\begingroup$ The group $G = \{e\}$ is abelian :) $\endgroup$
    – Igor Rivin
    Commented Apr 26, 2015 at 20:24
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    $\begingroup$ Igor, I do not get the joke. $\endgroup$
    – Tom Goodwillie
    Commented Apr 26, 2015 at 21:09
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    $\begingroup$ There was not any, I was confused. $\endgroup$
    – Igor Rivin
    Commented Apr 26, 2015 at 21:58

2 Answers 2

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Given a vector $u$ and an orthonormal basis $x_1,\ldots,x_n$, we have $||u||^2 = \left<u,x_1\right>^2 + \cdots + \left<u,x_n\right>^2$. But that means that, if you choose a nonzero $u$, then the function $F(x) = \left<u,x\right>^2$ gives a counterexample.

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    $\begingroup$ I sometimes wish MathOverflow was the kind of forum where one could just say "Great question! Ask your best undergraduates: they'll enjoy telling you how to do it more than I would." $\endgroup$
    – James Cranch
    Commented Apr 26, 2015 at 21:03
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    $\begingroup$ James, I hope you enjoyed it. It's such an attractive and edifying answer, too. $\endgroup$
    – Tom Goodwillie
    Commented Apr 26, 2015 at 21:09
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    $\begingroup$ Do these and constants generate all such functions? $\endgroup$
    – Douglas Zare
    Commented Apr 26, 2015 at 21:13
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    $\begingroup$ With the axiom of choice, there are lots of discontinuous automorphisms of $\mathbb{R}$. However, it might be feasible to classify the continuous functions $S^{n-1} \to \mathbb{R}$ satisfying the condition. Are they all in the closure of the linear combinations of constants and Cranch's functions? $\endgroup$
    – Douglas Zare
    Commented Apr 26, 2015 at 23:15
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    $\begingroup$ @DouglasZare For any such function, pullback by $O(n)$ gives another such function. So the space of such functions is a representation of $O(n)$, and because continuous functions decompose into spherical harmonics, this will decompose into different irreducible representations of $O(n)$. So you just have to test the different spherical harmonics. $\endgroup$
    – Will Sawin
    Commented Apr 27, 2015 at 1:09
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For the group $\mathbb{R}$, this was considered a long time ago by Andrew Gleason to solve a problem in the foundations of quantum mechanics.

http://www.iumj.indiana.edu/IUMJ/FULLTEXT/1957/6/56050

Such maps are called frame functions. For $n \geq 3$, frame functions $S^{n-1} \rightarrow \mathbb{R}$, taking positive values correspond to positive-definite operators on $\mathbb{C}^n$. When the sum is chosen to be 1, one gets density matrices.

Of course, if one drops the positivity and boundedness requirements, one can get horribly complicated functions using a nonlinear group automorphism of $\mathbb{R}$.

I am interested as to Tom Goodwillie's reasons for considering this problem.

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